How this code works?

Question

#include<stdio.h>
#include<conio.h>
void main()
{
char s[]="This is my character array";
int a=5;
printf(a>10?"%s":"%s",s);
}
Output:This is my character array

Answer 1

a>10? “%s” : “%s”
This is a ternary operator.

An expression a ? b : c evaluates to b if the value of a is true, and otherwise to c. One can read it aloud as "if a then b otherwise c".

In this code is a bit useless because it will print the string anyways.

Answer 2

Ur-Wellwisher yes this is correct u can use as many u want

Answer 3

According to standard your program will give you errors becz void main is no longer supportable in c cpp u can use int main and with return int u need to return any value at the end

Inside body u have define one char array and in print statement u have written a>10 if this will be ture then fist case will be execute otherwise second one will be execute this is ternary operator if u dont know about it u can read .
%s work as a format specifier in c language if u have to print any value then u need to use format specifiers for Integer value%d for char %c for string %s and simply s will be printed on your Console

Answer 4

Is this print statement is correct?
If so how many we can use it?